Naphthalene combustion may be used to calibrate the heat capacity of a bomb calorimeter. The heat of combustion of naphthalene is -40.1 kJ/g. When 0.8210 g of naphthalene was burned in a bomb calorimeter containing 1.000 kg of water, a temperature rise of 4.21°C was observed. What is the heat capacity of the bomb calorimeter excluding the water?

Question

Quizplus · Accepted Answer

The heat released by the combustion of naphthalene is absorbed by both the water and the calorimeter. The heat absorbed by the water can be calculated using the formula $q = mc\Delta T$, where $m$ is the mass of the water (1.000 kg or 1000 g), $c$ is the specific heat capacity of water (4.184 J/g°C), and $\Delta T$ is the temperature change (4.21°C). This gives $q_{water} = 1000 \times 4.184 \times 4.21 = 17615.44 \, \text{J} = 17.61544 \, \text{kJ}$.The total heat released by the naphthalene is its heat of combustion multiplied by the mass of naphthalene burned: $q_{naphthalene} = -40.1 \, \text{kJ/g} \times 0.8210 \, \text{g} = -32.9021 \, \text{kJ}$.The heat absorbed by the calorimeter is the difference between the total heat released and the heat absorbed by the water: $q_{calorimeter} = q_{naphthalene} - q_{water} = -32.9021 \, \text{kJ} - (-17.61544 \, \text{kJ}) = -15.28666 \, \text{kJ}$.Since the heat capacity ($C$) of the calorimeter is defined as the amount of heat required to raise its temperature by 1°C, and we know the calorimeter's temperature change is 4.21°C, we can rearrange the formula $q = C\Delta T$ to solve for $C$: $C = q_{calorimeter} / \Delta T = -15.28666 \, \text{kJ} / 4.21°C = -3.63 \, \text{kJ/°C}$.The negative sign indicates the direction of heat flow but is not needed when stating the heat capacity, so the correct answer is approximately 3.64 kJ/°C.

Naphthalene Combustion May Be Used to Calibrate the Heat Capacity